Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 340 / 2

Use the Fourier transform convention $\hat{\phi }(\xi )=\int \phi (x)e^{-ix\xi }dx$. Nesting and the orthonormal basis of $V_1$ give the scaling refinement equationHere $h_k=⟨\phi ,\sqrt{2}\phi (2\cdot -k)⟩$. Compact support makes these inner products zero for all but finitely many $k$, so $m$ is a trigonometric polynomial. Expanding the orthogonality of the integer translates of $\phi$ in the orthonormal basis of $V_1$ yieldsConsequentlyThis holds everywhere because $m$ is continuous. The invoked properties are nesting, dyadic dilation, orthonormal integer translates, and compact support; mere finite-energy refinement would not imply the quadrature mirror filter identity.

The density axiom and continuity of the Fourier transform at zero imply $|\hat{\phi }(0)|=1$. Here is a proof that avoids assuming a normalization of the integral. Choose a nonzero $f$ whose Fourier transform has bounded support. The MRA projection Fourier identity, with no aliasing once $j$ is sufficiently large, isSince $\phi \in L^1$, its Fourier transform is continuous and bounded, so this tends to $|\hat{\phi }(0){|}^2\Vert f{\Vert }_2^2$. Density and nesting give $P_{j}f\to f$ in $L^2$. Thus $|\hat{\phi }(0)|=1$, in particular it is nonzero. Evaluating the scaling refinement equation in frequency at zero now giveswhich is stronger than the requested absolute-value equality. Multiplying the scaling function by a constant of modulus one normalizes its integral to one without changing $m$.

At zero, the partial products are $m(0{)}^n$. Their assumed convergence and $|m(0)|=1$ force $m(0)=1$ and $\Theta (0)=1$. Uniform convergence on compact sets makes $\Theta$ continuous. The identity with $k=0$ gives $\Theta \in L^2$ with squared norm $2\pi$. Define $\phi$ by the inverse Fourier transform in $L^2$. The Plancherel theorem and the other given integral identities giveDefine $V_j$ as the closed span of ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$. These functions form an orthonormal basis of $V_j$, and dyadic dilation gives the scale axiom. Shifting the infinite product givesso the finite Fourier series of $m$ gives a scaling refinement equation and $V_j\subset V_{j+1}$. The coarse-scale projection argument of part 1(b) proves the trivial-intersection axiom.

For density, take a function $f$ with bounded Fourier support. The MRA projection Fourier identity gives, for large $j$,Uniform convergence of $\Theta (2^{-j}\xi )$ to one on that support proves the limit. Since $P_j$ is an orthogonal projection, $\Vert f-P_{j}f{\Vert }_2^2=\Vert f{\Vert }_2^2-\Vert P_{j}f{\Vert }_2^2\to 0$. Such functions are dense in $L^2$, givingThus all the multiresolution analysis axioms hold. Under the strong convergence and orthogonality assumptions supplied here, the additional nonvanishing condition is not needed in this last verification; it is useful when establishing those assumptions from a filter.

The factor $1+e^{-i\xi }$ has a simple zero at $\pi$, and $L(\pi )\ne 0$ makes the zero of $m$ there exactly order $N$. The quadrature mirror filter construction gives a high-pass symbol $m_1(\xi )=e^{-i\xi }\overline{m(\xi +\pi )}$, up to a constant phase, andThe normalized scaling function has $\hat{\phi }(0)=1$, so $\hat{\psi }$ has a zero of exactly order $N$ at zero. Compact support of the Daubechies wavelet permits differentiation under its Fourier transform:HenceThere are exactly $N$ vanishing moments, rather than just at least $N$.

Normalize $\hat{\phi }(0)=1$, as in the displayed product convention. Iterating the scaling refinement equation gives $\hat{\phi }(\xi )={\prod }_{j\ge 1}m(\xi /2^j)$. The finite-product identityhas the value one at zero by continuity. ThereforeWrite $M=\sup |L|$. Since $L(0)=1$, we have $M\ge 1$. For $|\xi |>1$, take $J=\lceil {\log }_2|\xi |\rceil$. The first $J$ factors are bounded by $M^J\le M|\xi {|}^{{\log }_{2}M}$. For the remaining factors, the trigonometric polynomial satisfies $|L(t)|\le 1+C|t|$ on $|t|\le 1$, henceUsing $|(1-e^{-i\xi })/(i\xi )|\le \min (1,2/|\xi |)$ givesThe strict hypothesis permits $\epsilon >0$ with $N-{\log }_{2}M\ge \alpha +1+\epsilon$. The given Fourier-decay criterion proves uniform Hölder regularity of exponent $\alpha$. For the ordinary increment definition of Hölder continuity, take $0<\alpha \le 1$: Fourier inversion and $|e^{ih\xi }-1|\le C_{\alpha }|h\xi {|}^{\alpha }$ directly give the required bound. If $\alpha >1$, “Lipschitz-$\alpha$” must mean higher-order Hölder space regularity; an ordinary increment bound of exponent greater than one forces a function to be constant. An unnormalized scaling function contributes its constant phase to the product formula.